Question

-1+sinAsin(90-A)/cot(90-A)=-sin^2A

Answer 1

i hope this helps i guess this is the only way

Answer 2

-1+[sinA·sin(90-A)/cot(90-A)]= -sin^2A
LHS
-1+[sinA·sin(90-A)/cot(90-A)]
-1+[sinA·cosA/tanA]                                {complementary angles}
-1+[(sinA·cosA)/(sinA/cosA)]                   {tanA=ainA/cosA}
-1+[(sinA·cosA) X (cosA/sinA)]                {sinA will get canceled}
-1+[cosA X cosA]
-1+cos^2A                                               {taking negative outside}
-[1-cos^2A]                                              {1-cos^2A=sin^2A}
-[sin^2A]
-sin^2A
Therefore LHS=RHS
So -1+[sinA·sin(90-A)/cot(90-A)]= -sin^2A